AP Calculus AB and BC

Integral of csc^2 x: Answer, Proof, and Steps

Q: What is the integral of csc^2 x?

It is -\cot x + C. The derivative of \cot x is -\csc^2 x, so reversing that rule and moving the minus sign onto the antiderivative gives -\cot x.

Q: Why is there a minus sign in the answer?

Because the derivative rule already carries one: \frac{d}{dx}\cot x = -\csc^2 x. To undo a derivative that produced -\csc^2 x you negate, so the antiderivative of +\csc^2 x is -\cot x.

Q: Is the integral of csc^2 x the same as the integral of sec^2 x?

No. \int \sec^2 x\,dx = \tan x + C with no minus sign, while \int \csc^2 x\,dx = -\cot x + C. The cosecant version carries the minus sign; the secant version does not.

The integral of csc^2 x with respect to x is -cot x + C. It is the direct companion of the derivative rule d/dx cot x = -csc^2 x, so reversing that rule gives the antiderivative. Differentiating -cot x returns csc^2 x, which confirms the result.

\int \csc^2 x\,dx = -\cot x + C

How to integrate csc^2 x

There is nothing to expand or substitute here. The integrand $\csc^2 x$ is exactly the derivative of a standard function with a sign attached, so the fastest route is to read the antiderivative straight off the trig derivative table.

The relevant derivative rule is the one for cotangent.

\frac{d}{dx}\cot x = -\csc^2 x

Reversing it, the function whose derivative is $\csc^2 x$ must be $-\cot x$ , since the minus sign in the derivative rule moves onto the antiderivative.

\int \csc^2 x\,dx = -\cot x + C

Check it by differentiating

Differentiate $-\cot x$ . The derivative of $\cot x$ is $-\csc^2 x$ , so the derivative of $-\cot x$ is $-(-\csc^2 x) = \csc^2 x$ , the original integrand. The two minus signs cancel, which is the whole reason the answer carries one.

Where the integral of csc^2 x shows up on the AP exam

Basic trig antiderivatives are part of Topic 6.8 in Unit 6 (Integration and Accumulation of Change), and Unit 6 carries 15 to 20 percent of the exam. This integral pairs with $\int \sec^2 x\,dx = \tan x + C$ , and the two are worth memorizing together because their signs behave oppositely.

\int_{\pi/4}^{\pi/2} \csc^2 x\,dx = \left[-\cot x\right]_{\pi/4}^{\pi/2} = 0 - (-1) = 1

Watch the domain on a definite integral. $\csc^2 x$ blows up wherever $\sin x = 0$ , so an interval that contains $x = 0$ or $x = \pi$ makes the integral improper rather than routine.

Common mistakes with the integral of csc^2 x

Dropping the minus sign and writing $\cot x + C$ . Differentiating $\cot x$ gives $-\csc^2 x$ , the wrong sign everywhere.
Confusing it with the secant case. $\int \sec^2 x\,dx = \tan x + C$ has no minus sign; the cosecant and cotangent pair carry the minus, the secant and tangent pair do not.
Answering $-\csc x$ . The antiderivative of $\csc^2 x$ is $-\cot x$ ; the form $-\csc x$ belongs to $\int \csc x\cot x\,dx$ .
Integrating across a vertical asymptote. On any interval containing a multiple of $\pi$ , $\csc^2 x$ is undefined and the integral does not converge.

Every answer on this page is machine checked

An automated test differentiates the antiderivative above and confirms it returns the integrand. A wrong sign or a missing factor fails the build, so it cannot reach you.

Frequently asked questions

What is the integral of csc^2 x?

It is $-\cot x + C$ . The derivative of $\cot x$ is $-\csc^2 x$ , so reversing that rule and moving the minus sign onto the antiderivative gives $-\cot x$ .

Why is there a minus sign in the answer?

Because the derivative rule already carries one: $\frac{d}{dx}\cot x = -\csc^2 x$ . To undo a derivative that produced $-\csc^2 x$ you negate, so the antiderivative of $+\csc^2 x$ is $-\cot x$ .

Is the integral of csc^2 x the same as the integral of sec^2 x?

No. $\int \sec^2 x\,dx = \tan x + C$ with no minus sign, while $\int \csc^2 x\,dx = -\cot x + C$ . The cosecant version carries the minus sign; the secant version does not.